命名查询的等效条件查询

发布于 2024-12-12 15:11:29 字数 586 浏览 0 评论 0原文

我的命名查询看起来像这样，感谢这里.

@NamedQuery(
name="Cat.favourites", 
query="select c 
      from Usercat as uc 
      inner join uc.cat as c 
      where uc.isFavourtie = true 
      and uc.user = :user")

对实现的调用如下所示：

Session session = sessionFactory.getCurrentSession();
Query query = session.getNamedQuery("Cat.favourites");
query.setEntity("user", myCurrentUser);
return query.list();

返回猫列表的等效条件查询是什么？

原文

My named query looks like this, thanks to here.

@NamedQuery(
name="Cat.favourites", 
query="select c 
      from Usercat as uc 
      inner join uc.cat as c 
      where uc.isFavourtie = true 
      and uc.user = :user")

And the call to implement looks like this :

Session session = sessionFactory.getCurrentSession();
Query query = session.getNamedQuery("Cat.favourites");
query.setEntity("user", myCurrentUser);
return query.list();

What would be the equivalent criteria query that returns a list of cats ?

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灵芸 2024-12-19 15:11:30

根据 JPA 2.0 标准：
（这是使用 JPA 2.0 Criteria api 实现此目的的多种方法之一）

final CriteriaQuery<Cat> cq = getCriteriaBuilder().createQuery(Cat.class);
final CriteriaBuilder cb = entityManager.getCriteriaBuilder();

final Root<Usercat> uc= cq.from(Usercat.class);

cq.select(uc.get("cat");

Predicate p = cb.equal(uc.get("favourtie", true);
p = cb.and(p, cb.equal(uc.get("user"), user));
cq.where(p);

final TypedQuery<Cat> typedQuery = entityManager.createQuery(cq);
return typedQuery.getResultList();

With JPA 2.0 Criteria:
(This is one of the many ways you can achieve this using JPA 2.0 Criteria api)

final CriteriaQuery<Cat> cq = getCriteriaBuilder().createQuery(Cat.class);
final CriteriaBuilder cb = entityManager.getCriteriaBuilder();

final Root<Usercat> uc= cq.from(Usercat.class);

cq.select(uc.get("cat");

Predicate p = cb.equal(uc.get("favourtie", true);
p = cb.and(p, cb.equal(uc.get("user"), user));
cq.where(p);

final TypedQuery<Cat> typedQuery = entityManager.createQuery(cq);
return typedQuery.getResultList();

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